A charge of #40 C# passes through a circuit every #8 s#. If the circuit can generate #42 W# of power, what is the circuit's resistance?

2 Answers
Feb 10, 2016

#R = 1.68 Omega#

Explanation:

The current, #I#, is the charge passing through per unit time (#Q/C#).

So,

#I = frac{40"C"}{8"s"} = 5"A"#

The power arises from the fact that charges are moving over a potential difference.

#P=IV#

From Ohm's Law, we know that #V=IR# for an ideal resistance.

Therefore,

#P = I^2R#

#42"W" = (5"A")^2R#

#R = 1.68 Omega#

Feb 10, 2016

#1.68 Omega#

Explanation:

We'll need 2 formulae to answer this. namely:

#Q=It# and #P = I^2R#

Where Q is charge (40C), t is time (8s), P is power (42W), I is current and R is resistance.

Begin by using the first formula to obtain the current:

#I = Q/t = 40/8 = 5A#

Now use the second formula to get the resistance:

#R = P/I^2=42/5^2=42/25=1.68 Omega#